A 5 percent aqueous solution of …
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Sia ? 6 years, 7 months ago
Given,5% sugar solution means, {tex}w_B{/tex} = 5g
{tex}{w_A} = 95g,\,{M_B} = 342g\,mo{l^{ - 1}}{/tex}
Similarly, for 5% glucose solution, {tex}{w_B}(glu\cos e) = 5g{/tex}
{tex}{w_A} = 95\,g\,and {M_B}(glu\cos e) = 180\,g\,mo{l^{ - 1}}{/tex}
{tex}\Delta {T_f}(cane\,sugar) = \,273.15K - 271K = 2.15K{/tex}
{tex}\Delta {T_f} = \frac{{{K_f} \times {w_B} \times 1000}}{{{M_B}{w_A}}}{/tex}
{tex}2.15 = \frac{{{K_f} \times 5 \times 1000}}{{342 \times 95}}{/tex}
{tex}{k_f} = \frac{{2.15 \times 342 \times 95}}{{5 \times 1000}} = 13.9707{/tex}
similarly, for glucose
{tex}\Delta {T_f}(glu\cos e) = \frac{{13.9707 \times 5 \times 1000}}{{180 \times 95}}{/tex}
{tex}\Delta {T_f} = 4.085K{/tex}
Freezing point of the solution
= 273.15K - 4.085K = 269.065 K
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